Problem: You know, it's hard to conduct a show with lots of participants and spectators at the same place nowadays. Still, you are not giving up on your dream to make a car crash showcase! You decided to replace the real cars with remote controlled ones, call the event "Remote Control Kaboom Show" and stream everything online.

For the preparation you arranged an arena — an infinite 2D-field. You also bought n remote controlled cars and set them up on the arena. Unfortunately, the cars you bought can only go forward without turning left, right or around. So you additionally put the cars in the direction you want them to go.

To be formal, for each car i (1 ≤ i ≤ n) you chose its initial position (x_i, y_i) and a direction vector (dx_i, dy_i). Moreover, each car has a constant speed s_i units per second. So after car i is launched, it stars moving from (x_i, y_i) in the direction (dx_i, dy_i) with constant speed s_i.

The goal of the show is to create a car collision as fast as possible! You noted that launching every car at the beginning of the show often fails to produce any collisions at all. Thus, you plan to launch the i-th car at some moment t_i. You haven't chosen t_i, that's yet to be decided. Note that it's not necessary for t_i to be integer and t_i is allowed to be equal to t_j for any i, j.

The show starts at time 0. The show ends when two cars i and j (i ≠ j) collide (i. e. come to the same coordinate at the same time). The duration of the show is the time between the start and the end.

What's the fastest crash you can arrange by choosing all t_i? If it's possible to arrange a crash then print the shortest possible duration of the show. Otherwise, report that it's impossible.

Input

The first line contains a single integer n (1 ≤ n ≤ 25000) — the number of cars.

Each of the next n lines contains five integers x_i, y_i, dx_i, dy_i, s_i (-10^3 ≤ x_i, y_i ≤ 10^3; 1 ≤ |dx_i| ≤ 10^3; 1 ≤ |dy_i| ≤ 10^3; 1 ≤ s_i ≤ 10^3) — the initial position of the i-th car, its direction vector and its speed, respectively.

It's guaranteed that all cars start at distinct positions (i. e. (x_i, y_i) ≠ (x_j, y_j) for i ≠ j).

Output

Print the shortest possible duration of the show if it's possible to arrange a crash by choosing all t_i. Otherwise, print "No show :(".

Your answer is considered correct if its absolute or relative error does not exceed 10^{-6}.

Formally, let your answer be a, and the jury's answer be b. Your answer is accepted if and only if \frac{|a - b|}{max{(1, |b|)}} ≤ 10^{-6}.

Examples

Input


4
3 -1 -1 1 2
2 3 -3 -2 10
-4 2 1 -2 1
-2 -2 -1 2 4


Output


0.585902082262898


Input


2
-1 1 -1 1 200
1 1 1 5 200


Output


No show :(

Note

Here is the picture for the first example: 

<image>

The fastest cars to crash are cars 2 and 4. Let's launch car 2 at 0, car 4 at about 0.096762 and cars 1 and 3 at arbitrary time. That way cars 2 and 4 will crash into each other at about 0.585902. So here's what it looks like at the moment of the collision:

<image>

Here's the picture for the second example:

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast", "unroll-loops", "omit-frame-pointer", "inline")
#pragma GCC option("arch=native", "tune=native", "no-zero-upper")
#pragma GCC target("avx2")
using namespace std;
template <typename T>
void maxtt(T &t1, T t2) {
  t1 = max(t1, t2);
}
template <typename T>
void mintt(T &t1, T t2) {
  t1 = min(t1, t2);
}
bool debug = 0;
int n, m, k;
string direc = "RDLU";
const long long MOD2 = (long long)998244353 * (long long)998244353;
long long ln, lk, lm;
void etp(bool f = 0) {
  puts(f ? "YES" : "NO");
  exit(0);
}
void addmod(int &x, int y, int mod = 998244353) {
  x += y;
  if (x >= mod) x -= mod;
  if (x < 0) x += mod;
  assert(x >= 0 && x < mod);
}
void et(int x = -1) {
  printf("%d\n", x);
  exit(0);
}
long long fastPow(long long x, long long y, int mod = 998244353) {
  long long ans = 1;
  while (y > 0) {
    if (y & 1) ans = (x * ans) % mod;
    x = x * x % mod;
    y >>= 1;
  }
  return ans;
}
long long gcd1(long long x, long long y) { return y ? gcd1(y, x % y) : x; }
int sign(long double k) {
  if (k > 1e-9)
    return 1;
  else if (k < -1e-9)
    return -1;
  return 0;
}
struct Point {
  long double x, y;
  Point() {}
  Point(long double x, long double y) : x(x), y(y) {}
  long double abs() const { return hypot(x, y); }
  long double arg() const { return atan2(y, x); }
  Point operator*(double o) const { return Point(x * o, y * o); }
  Point operator+(const Point &o) const { return Point(x + o.x, y + o.y); }
  Point operator-(const Point &o) const { return Point(x - o.x, y - o.y); }
  Point operator/(double o) const { return Point(x / o, y / o); }
  bool operator<(const Point &o) const {
    return x < o.x - 1e-9 || (x < o.x + 1e-9 && y < o.y - 1e-9);
  }
  friend bool operator==(const Point &r1, const Point &r2) {
    return r1.x == r2.x && r1.y == r2.y;
  }
  Point scale(double o) const { return *this * (o / abs()); }
  Point rotY() const { return Point(-y, x); }
  Point rotX() const { return Point(y, -x); }
  long double cross(Point b) const { return x * b.y - b.x * y; }
  long double dot(Point b) const { return x * b.x + y * b.y; }
  long double disToSeg(Point &a, Point &b) {
    return fabs(((*this) - a).cross(b - a) / (b - a).abs());
  }
  pair<long long, long long> getVec() {
    if (x == 0) return {0, y > 0 ? 1 : -1};
    if (y == 0) return {x < 0 ? -1 : 1, 0};
    long long gg = gcd1(::abs(x), ::abs(y));
    long long px = x / gg, py = y / gg;
    return {px, py};
  }
  void readin() { scanf("%lf%lf", &x, &y); }
};
struct Line {
  Point s, t;
  int id;
  Line() {}
  Line(Point p, Point q) {
    s = p;
    t = q;
  }
  long double get_y(long double x) const {
    if (sign(s.x - t.x) == 0) return s.y;
    return s.y + (t.y - s.y) * (x - s.x) / (t.x - s.x);
  }
};
int cmp(long double k1, long double k2) { return sign(k1 - k2); }
int intersect(long double l1, long double r1, long double l2, long double r2) {
  if (l1 > r1) swap(l1, r1);
  if (l2 > r2) swap(l2, r2);
  return cmp(r1, l2) != -1 && cmp(r2, l1) != -1;
}
int checkSS(Point k1, Point k2, Point k3, Point k4) {
  return intersect(k1.x, k2.x, k3.x, k4.x) &&
         intersect(k1.y, k2.y, k3.y, k4.y) &&
         sign((k3 - k1).cross(k4 - k1)) * sign((k3 - k2).cross(k4 - k2)) <= 0 &&
         sign((k1 - k3).cross(k2 - k3)) * sign((k1 - k4).cross(k2 - k4)) <= 0;
}
int intersect(Line a, Line b) { return checkSS(a.s, a.t, b.s, b.t); }
bool operator<(const Line &a, const Line &b) {
  long double x = max(min(a.s.x, a.t.x), min(b.s.x, b.t.x));
  return sign(a.get_y(x) - b.get_y(x)) <= 0;
}
struct has_inter {
  struct event {
    long double x;
    int tp, id;
    event() {}
    event(long double x, int tp, int id) : x(x), tp(tp), id(id) {}
    bool operator<(const event &e) const {
      if (sign(x - e.x) == 0) return tp > e.tp;
      return sign(x - e.x) < 0;
    }
  };
  set<Line> s;
  vector<set<Line>::iterator> where;
  inline set<Line>::iterator prev(set<Line>::iterator it) {
    return it == s.begin() ? s.end() : --it;
  }
  inline set<Line>::iterator next(set<Line>::iterator it) { return ++it; }
  pair<int, int> solve(vector<Line> &a) {
    int n = a.size();
    vector<event> e;
    for (int i = 0; i < n; i++) {
      e.push_back(event(min(a[i].s.x, a[i].t.x), +1, i));
      e.push_back(event(max(a[i].s.x, a[i].t.x), -1, i));
    }
    sort(e.begin(), e.end());
    s.clear();
    where.resize(a.size());
    for (int i = 0; i < e.size(); i++) {
      int id = e[i].id;
      if (e[i].tp == +1) {
        auto nxt = s.lower_bound(a[id]);
        auto prv = prev(nxt);
        if (nxt != s.end() && intersect(*nxt, a[id]))
          return make_pair(nxt->id, id);
        if (prv != s.end() && intersect(*prv, a[id]))
          return make_pair(prv->id, id);
        where[id] = s.insert(nxt, a[id]);
      } else {
        if (where[id] == s.end()) continue;
        auto nxt = next(where[id]), prv = prev(where[id]);
        if (nxt != s.end() && prv != s.end() && intersect(*nxt, *prv))
          return make_pair(prv->id, nxt->id);
        s.erase(where[id]);
      }
    }
    return make_pair(-1, -1);
  }
};
int x[25135], y[25135], dx[25135], dy[25135], s[25135];
bool ck(long double t) {
  vector<Line> vp;
  for (int(i) = 1; (i) <= (int)(n); (i)++) {
    long double d = sqrt(dx[i] * dx[i] + dy[i] * dy[i]);
    Point k1 = Point(x[i], y[i]);
    Point k2 = Point(x[i] + t * s[i] * dx[i] / d, y[i] + t * s[i] * dy[i] / d);
    Line l = {k1, k2};
    l.id = i;
    vp.push_back(l);
  }
  has_inter Q;
  return Q.solve(vp).first != -1;
}
void fmain(int tid) {
  scanf("%d", &n);
  for (int(i) = 1; (i) <= (int)(n); (i)++)
    scanf("%d%d%d%d%d", x + i, y + i, dx + i, dy + i, s + i);
  long double l = 0, r = 1e10;
  bool ok = 0;
  for (int(i) = 0; (i) < (int)(100); (i)++) {
    long double mid = (l + r) / 2;
    if (ck(mid)) {
      ok = 1;
      r = mid;
    } else
      l = mid;
  }
  if (!ok)
    puts("No show :(");
  else
    printf("%.10Lf\n", l);
}
int main() {
  int t = 1;
  for (int(i) = 1; (i) <= (int)(t); (i)++) {
    fmain(i);
  }
  return 0;
}